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Math Help - Baby Rudin - ch 2, ex 27

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    Baby Rudin - ch 2, ex 27

    This is a question regarding exercise 27 in chapter 2 from baby Rudin:

    How does one prove that P is perfect?
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    Quote Originally Posted by jefferson_lc View Post
    This is a question regarding exercise 27 in chapter 2 from baby Rudin:

    How does one prove that P is perfect?
    It would be helpful if you posted the full question. Most people don't have a stack of old textbooks with them at all times.
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    That depends on two things: (1) What is the definition of P? (2) What is the definition of "perfect"?
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    Rhymes with Orange Chris L T521's Avatar
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    Here's the problem from Rudin's "Principles of Mathematical Analysis":

    Quote Originally Posted by Chapter 2, Exercise 27
    Define a point p in a metric space X to be a condensation point of a set E\subset X if every neighborhood of p contains uncountably many points of E.

    Suppose E\subset \mathbb{R}^k, E is countable, and let P be the set of all condensation points of E. Prove that P is perfect and that at most countably many points of E are not in P. In other words, show that P^c\cap E is at most countable. Hint: Let \{V_n\} be a countable base for \mathbb{R}^k, let W be the union of those V_n for which E\cap V_n is at most countable, and prove P=W^c.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Here's the problem from Rudin's "Principles of Mathematical Analysis":
    It should be uncountable, otherwise it's trivially true that P is perfect since it' empty!

    Lemma: Let X be a second countable metric (or topological) space with countable basis \mathcal{B} and E\subseteq X. Then, if \Omega=\{U_\alpha\}_{\alpha\in\mathcal{A}} is an open cover for E then \Omega admits a countable subcover.
    Proof: For each x\in E there exists a neighborhood N_x containing it. But, since \mathcal{B} is a basis we may find some basic open set B_x\in\mathcal{B} for which x\in B_x\subseteq N_x. Evidently then \left\{B_x:x\in E\right\} is an open cover for E, and since it's a subset of \mathcal{B} it must be countable. Thus, for each element B_{x_0} of \left\{B_x:x\in E\right\} choosing some N_{x_0}\in\Omega for which B_{x_0}\subseteq N_{x_0} produces a countable subcover for \Omega as desired. \blacksquare

    So, we now claim that P, as described in the problem, is perfect. To sync our notations call D(P) (the derived set) the set of all limit points of P. Thus, we aim to show that D(P)=P.

    So, to first see that D(P)\subseteq P let x\in D(P). Then, for each neighborhood U of x we have that there is some y\in D(P)\cap U. But, since U is a neighborhood of y we see that U\cap E is uncountable. Thus, since U was arbitrary it follows that x\in P.

    Conversely, suppose that x\notin D(P) then, there exists a neighborhood U of x for which every element y of U-\{x\} has a neighborhood N_y for which \#\left(N_y\cap E\right)\leqslant\aleph_0 clearly then \left\{N_y\right\}_{y\in U-\{x\}} is an open cover for U-\{x\}. Thus, by our lemma it must admit a countable subcover \left\{N_{y_n}\}_{n\in\mathbb{N}}. Note though that \displaystyle \#\left(\left(U-\{x\}\cap E\right)\right)=\#\left(\bigcup_{n\in\mathbb{N}}N_  {y_n}\cap E\right)=\#\left(\bigcup_{n\in\mathbb{N}}\left(N_{  y_n}\cap E\right)\right)\leqslant \aleph_0 (this last part since by construction N_{y_n}\cap E is countable and the countable union of countable sets is countable). and thus \#\left(U\cap E\right)\leqslant\aleph_0 and thus x\notin P. Thus, \mathbb{R}^n-D(P)\subseteq \mathbb{R}^n-P\Leftrightarrow P\subseteq D(P).

    Thus, putting these two together gives D(P)=P.


    So, now for the second part of the problem let \mathcal{B} be the countable basis for \mathbb{R}^n (the set of all open balls with rational radii at points of \mathbb{Q}^n) and let \mathcal{W}=\left\{B\in\mathcal{B}:\#\left(B\cap E\right)\leqslant\aleph_0\right\}. We claim that \displaystyle \bigcup_{B\in\mathcal{W}}B=\mathbb{R}-P. To see this let \displaystyle x\in\bigcup_{B\in\mathcal{W}}B, then x\in B_0 for some B_0\in\mathcal{W}, but evidently then B_0 is a neighborhood of x for which \#\left(B_0\cap E\right)\leqslant\aleph_0 and so x\notin P. Conversely, let x\notin P, then there is some neighborhood U of x for which \#\left(U\cap E\right)\leqslant\aleph_0, and since \mathcal{B} is a basis we man find some B_0\in\mathcal{B} for which x\in B_0\subseteq U and so evidently \#\left(B_0\cap E\right)\leqslant\aleph_0 and so B_0\in\mathcal{W} and thus \displaystyle x\in B_0\subseteq\bigcup_{B\in\mathcal{W}}B. It follows that \displaystyle \mathbb{R}^n-P=\bigcup_{B\in\mathcal{W}}B as desired. Note though that \displaystyle \#\left(E\cap \left(\mathbb{R}^n-P\right)\right)=#\left(E\cap\bigcup_{B\in\mathcal{  W}}B\right)=\#\left(\bigcup_{B\in\mathcal{W}}\left  (B\cap E\right)\right)\leqslant \aleph_0 as desired (this past part gotten since by construction \#\left(B\cap E\right)\leqslant\aleph_0 for each B\in\mathcal{W} and since \mathcal{W}\subseteq\mathcal{B} we have that \mathcal{W} is countable).
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